Engineering Mathematics GATE

Test Paper-1Engineering Mathematics

Source Book: GATE ENGINEERING MATHEMATICS

Author: RK Kanodia

ISBN: 9788192276267

Publisher : Nodia and Company

Visit us at: www.nodia.co.in

Q. No. 1 - 10 Carry One Mark Each

MCQ 1.1 Every diagonal elements of a Hermitian matrix is(A) Purely real (B) 0

(C) Purely imaginary (D) 1

SOL 1.1 A square matrix A is said to Hermitian if A AQ = . So a aij ji= . If i j= then a aii ii= i.e. conjugate of an element is the element itself and aii is purely real.Hence (A) is correct option.

MCQ 1.2 If A is a 3-rowed square matrix, then adj(adj A) is equal to(A) A 6 (B) A 3

(C) A 4 (D) A 2

SOL 1.2 We have ( )adj adjA A ( )n 1 2

= −

Putting 3n = , we get ( )adj adjA A 4=Hence (C) is correct option.

MCQ 1.3 If z xyf xy= a k, then x

xz y

yz

22

22+ is equal to

(A) z (B) 2z

(C) xz (D) yz

SOL 1.3 The given function is homogeneous of degree 2.

Euler’s theorem xxz y

yz

22

22+ 2z=

Hence (B) is correct option.

MCQ 1.4 ( ) ( )e f x f x dxx + l" ,# is equal to(A) ( )e f xx l (B) ( )e f xx

(C) ( )e f xx + (D) None of these

SOL 1.4

Let I ( ) ( )e f x f x dxx= + l" ,#

Page 2 Engineering Mathematics Test Paper-1

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( ) ( )e f x dx e f x dxx x= + l##

( ) ( ) ( ) ( ) .f x e f x e dx e f x dx f x ex x x x= − + =l l& 0# #


MCQ 1.5 The solution of the differential equation 2 2xdxdy y= − is

(A) 2y xc= − (B) 2y x

c= +

(C) 2y c x= − (D) 2y c x= +

SOL 1.5 This is separable and written as ydy

xdx

22− =

Integrating 2 (2 )ln y− − ln lnx c= −

& (2 )lnc y 2− − ln 2x y xc

&= = −

Hence (A) is correct option.

MCQ 1.6 If 2v xy= , then the analytic function ( )f z u iv= + is (A) z c2 + (B) z c2 +−

(C) z c3 + (D) z c3 +−

SOL 1.6

xv

22 2 ( , ), 2 ( , )y h x y

yv x g x y

22= = = =

by Milne’s Method ( )f zl ( ,0) ( ,0) 2 0 2g z ih z z i z= + = + =On integrating ( )f z z c2= +Hence (A) is correct option.

MCQ 1.7 The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is(A) 5 (B) 7

(C) 8 (D) 10

SOL 1.7 Since 8 occurs most often, mode 8=Hence (C) is correct option.

MCQ 1.8 If the probabilities that A and B will die within a year are p and q respectively, then the probability that only one of them will be alive at the end of the year is(A) pq (B) (1 )p q−

(C) (1 )q p− (D) 2p q pq+ −

SOL 1.8 Required probability P= [(A dies and B is alive) or (A is alive and B dies)]


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(1 ) (1 ) 2p q p q p q pq= − + − = + −Hence (D) is correct option.

MCQ 1.9 If cov( , ) 10X Y = , var( ) 6.25X = and var( ) 31.36Y = , then ( , )X Yρ is

(A) 75 (B) 5

4

(C) 43 (D) 0.256

SOL 1.9

( , )X Yρ( ) ( )

( , ). .

covX Y

X Y6 25 31 36

1075

var var #= = =


MCQ 1.10 Let r be the correlation coefficient between x and y and byx , bxy be the regression coefficients of y on x and x on y respectively then(A) r b bxy yx= + (B) r b bxy yx#=

(C) r b bxy yx#= (D) ( )r b b21

xy yx= +

SOL 1.10

.b r xy

yx σσ= and .b r y

xxy σ

σ=

r2 b b r b bxy yx xy yx& ##= =Hence (C) is correct option.

Q. No. 11- 21 Carry Two Mark Each

MCQ 1.11 The root of equation 2 7logx x10− = by regular false method correct to three places of decimal is(A) 3.683 (B) .3 789

(C) 3.788 (D) .3 790

SOL 1.11 Let ( )f x 2 7logx x10= − −Taking 3.5, 4x x0 1= = , in the method of false position, we get

x2 ( ) ( )( ) 3.5 . .

. ( 0.5441)xf x f x

x x f x 0 3979 0 54410 5

01 0

1 00= − −

− = − + −

3.7888=Since (3.7888) 0.0009f =− and (4) 0.3979f = , therefore the root lies between 3.7888 and 4.Taking 3.7888, 4x x0 1= = , we obtain

x3 3.7888 .. ( .009) 3.78930 3988

0 2112= − − =

Hence the required root correct to three places of decimal is 3.789.


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MCQ 1.12 For the differential equation x ydxdy 2= − given that

:x 0 0.2 0.4 0.6

:y 0 0.02 0.0795 0.1762

Using Milne predictor-correction method, the y at next value of x is(A) 0.2498 (B) 0.3046

(C) 0.4648 (D) 0.5114

SOL 1.12 :x 0 0.2 0.4 0.6

On calculation we get ( )f x x y2= − ( )f x1 .0 1996= ( )f x2 .0 3937= ( )f x3 .0 5689=Using predictor formula

y ( )p4 (2 2 )y h f f f3

40 1 2 3= + − +

Here 0.2h =

y ( )p4 . [ (. ) (. ) (. )]0

30 8 2 1996 3937 2 5689= + − +

y ( )c4 ( 4 ),y h f f f

3*

2 2 3 3= + − +

f *4 ( , ) (. , . ) .f x y f 8 0 3049 07070( )p

4 4= = =

y ( )c4 . [. (. ) . ] .0795

302 3937 4 5689 7070 3046= + + + =


MCQ 1.13 The ranks obtained by 10 students in Mathematics and Physics in a class test are a follows

Rank in Maths Rank in Chem.

1 3

2 10

3 5

4 1

5 2

6 9


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7 4

8 8

9 7

10 6

The coefficient of correlation between their ranks is(A) 0.15 (B) 0.224

(C) 0.625 (D) None

SOL 1.13

r ( ) . ( )

( , ).

. 0.97cov

varx yx y

2 89 10016 5

var #= = − =−

Hence (D) is correct option.

MCQ 1.14 A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve a problem, selected at random from the book?(A) 0.16 (B) 0.63

(C) 0.97 (D) 0.20

SOL 1.14 Let E = the event that A solves the problem, andF = the event that B solves the problem.Clearly E and F are independent events.

( )P E 0.910090= = ,

( )P F 0.710070= = ,

( )P E F/ ( ) . ( ) 0.9 0.7 0.63P E P F #= = =Required probability ( )P E F,= ( ) ( ) ( )P E P F P E F/= + − (0.9 0.7 0.63) 0.97= + − =Hence (C) is correct option.

MCQ 1.15 ( )( )

?x a x b

x dx2 2 2 2

2

+ +=

3

3

−#

(A) a babπ+ (B)

( )aba bπ +

(C) a bπ+ (D) ( )a bπ +

SOL 1.15


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I ( )( )

( )z a z b

z dz f z dzcc

2 2 2 2

2

=+ +

= ##

where c is be semi circle r with segment on real axis from R− to R .The poles are ,z ia z ib! != = . Here only z ia= and z ib= lie within the contour c

( )f z dzc

# 2 iπ=

(sum of residuces at z ia= and z ib= )Residue at z ia= , is

( )( )( )( ) ( )

lim z ibz ia z ia z b

zi a b

a2z b 2 2

2

2 2= −− − +

=−"

Residuce at z ib= is

( )( )( )( )( )

lim z ibz ib z ia z ib z ib

zz b

2

= − − + + −"

( )i a b

b2 2 2=

−−

( )f z dzc

# ( ) ( )f z dz f z dzR

R

r

= +−##

( )2 ( )

i a bi a b a b2 2 2

π π=−

− = +

Now ( )f z dzr

# ( )( )

ReR e a R e b

ie i di i

i

2 2 2 2 2 2

2

0

θ=+ +

π

θ θ

θ ιθ

#

e

Ra e

Rb

Re d

i i

i

22

22

2

2

3

0

θ=

+ +

π

θ θ

θ

c cm m

#

Now when , ( )R b z dz 0r

" 3 =#

( )( )x a x bx dz a b2 2 2 2

2 π+ +

= +3

3

−#

Hence (C) is correct option.

MCQ 1.16 ?z e dz/z

c

2 1 =# where c is z 1=

(A) 3i π (B) 3i π−

(C) i3π (D) None of the above

SOL 1.16

( )f z ! !

...z e z z z z1 1

21

31z2 1 2

2 3= = + + + +b l

...z z z21

612= + + + +


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The only pole of ( )f z is at 0z = , which lies within the circle 1z = ( )f z dz

c

# 2 iπ= (residue at 0z = )

Now, residue of ( )f z at 0z = is the coefficient of z1 i.e. 6

1

( )f z dzc

# 2 i i61

31π π#= =


MCQ 1.17 The family of conic represented by the solution of the DE (4 3 1) (3 2 1)x y dx x y dy+ + + + + 0= is(A) Circles (B) Parabolas

(C) Hyperbolas (D) Ellipses

SOL 1.17 In the given equation 4 3 1M x y= + + and 3 2 1N x y= + + ,

yM

22 3, 3

xN

22= =

Thus yM

22 ,

xN

22=

Hence the given equation is exact. The solution is

( ) ( )x y dx y dy4 3 1 2 1+ + + +## c=

2 3x xy x y2 2+ + + c= ...(i)

Here, 2, 1,a b h= = 23=

h ab2 − 2 1 049

41 >#= − =

Hence, (i) represents a family of hyperbolas.Hence (C) is correct option.

MCQ 1.18 The equation of the curve, for which the angle between the tangent and the radius vector is twice the vectorial angle is 2sinr A2 θ= . This satisfies the differential equation

(A) 2tanrddr

θ θ= (B) 2tanrdrdθ θ=

(C) 2cosr ddr

θ θ= (D) 2cosr drdθ θ=

SOL 1.18 Given that r 2 2sinA θ= ...(i)Differentiating (i) with respect to θ, we get

2rddr

θ 2 2cosA θ=

or, rddr

θ 2cosA θ= ...(ii)


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Eliminating ‘A’ between (i) and (ii), we get

r ddr

θ 2sin cosr2

2

θ θ=

or, r drdθ 2tan θ=

Which is the required differential equation.Hence (B) is correct option.

MCQ 1.19 If ( ) , 1logF a a a1 >= and ( )F x a dx Kx= +# is equal to

(A) ( 1)loga a a1 x a− + (B) ( )loga a a1 x a−

(C) ( 1)loga a a1 x a+ + (D) ( 1)loga a a1 x a+ −

SOL 1.19

( )F x loga dx K aa Kx

x

= + = +#

& ( )F a logaa K

a

= +

K log log loga aa

aa1 1a a

= − = −

( )F x [ ]log log logaa

aa

a a a1 1 1x n

x a= + − = − +


MCQ 1.20 For what value of x x0 2# # πa k, the function

( )tany

xx

1= + has a maxima ?

(A) tanx (B) 0

(C) cotx (D) cosx

SOL 1.20

Let z tan tanx

xx x

x1 1= + = +

Then, dxdz sec

xx1

22=− + and 2 sec tan

dxd z

xx x2

2

2

32= +

dxdz 0 0sec cos

xx x x1

22& &= − + = =

dxd z

cosx x2

2

=; E 2 2 0cos sec tanx x x >3 2= +

Thus z has a minima and therefore y has a maxima at cosx x= .Hence (D) is correct option.

MCQ 1.21 If cossin

sincosA

αα

αα= −α > H, then consider the following statements :


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I. A A A: =α β βα II. A A A( ): =α β α β+

III. ( )cossin

sincos

A nn

n

n

n

αα

αα= −α > H IV. ( )

cossin

sincos

nn

nnA n

αα

αα= −α > H

Which of the above statements are true ?(A) I and II (B) I and IV

(C) II and III (D) II and IV

SOL 1.21

A A:α β cossin

sincos

cossin

sincos

αα

αα

ββ

ββ= − −> >H H

cossin

sincos

nn

nn

αα

αα= −> H A= α β+

Also, it is easy to prove by induction that

( )A nα

cossin

sincos

nn

nn

αα

αα= −> H


Common data for Question 22 to Q. 23

( )f z0 ( )z z

z z dz3 7 1

c0

2

= −+ +# , where c is the circle 4x y2 2+ =

MCQ 1.22 The value of (3)f is(A) 6 (B) 4i

(C) 4i− (D) 0

SOL 1.22

(3)f zz z dz3

3 7 1

c

2

= −+ +# , since 3z0 = is the only singular

point of zz z dz3

3 7 12

−+ + , and it lies outside the circle 4x y2 2+ = i.e.,

2z = ,therefore zz z

33 7 12

−+ + is analytic everywhere within c .

Hence by Cauchy’s theorem -

(3)f 0zz z dz3

3 7 1

c

2

= −+ + =#


MCQ 1.23 The value of (1 )f i−l is(A) 7( 2)iπ + (B) 6(2 )iπ+

(C) 2 (5 13)iπ + (D) 0

SOL 1.23 The point (1 )i− lies within circle 2z = (... the distance of 1 i− i.e., (1,1) from


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the origin is 2 which is less than 2, the radius of the circle).Let ( ) 3 7 1z z z2φ = + + then by Cauchy’s integral formula

z zz z dz3 7 1

c0

2

−+ +# 2 ( )i z0π φ=

& ( )f z0 2 ( )i z0π φ= ( ) 2 ( )f z i z0 0& π φ=l l

and ( )f z0m 2 ( )i z0π φ= m

since, ( )zφ 3 7 1z z2= + +& ( )zφl 6 7z= + and ( ) 6zφ =m

(1 )f i−l 2 [6(1 ) 7] 2 (5 13 )i i iπ π= − + = +Hence (C) is correct option.

Common data for Question 24 to Q. 25

Expand the function ( )( )z z1 2

1− − in Laurent’s series for the condition given in

question.

MCQ 1.24 1 2z< <

(A) ...z z z1 2 3

2 3+ + +

(B) ... ...z z z z z z21

41

81

1813 2 1 2 3− − − − − − − −− − −

(C) ...z z z1 3 72 2 4+ +

(D) None of the above

SOL 1.24

Here ( )f z ( )( )z z z z1 2

12

11

1= − − = − − − ...(i)

Since, 1z > 1z1 <& and 2 1z

z2< <&

z 11−

z zz z1 1

1 1 1 1 1=

−= −

−

bb

ll

and z 21− ...z z z z

21 1 2 2

1 1 2 4 91 2 3

= − − =− + + + +−a k : D

equation (1) gives -

( )f z ... ...z z zz z z z2

1 1 2 4 91 1 1 1 12 3

2 3=− + + + + − + + + +b bl l

or ( )f z ... ...z z z z z z21

41

81

1813 2 1 2 3= − − − − − − − −− − −



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MCQ 1.25 2z >

(A) ...z z z6 13 20

2 3+ + + (B) ...z z z1 8 13

2 3+ + +

(C) ...z z z1 3 72 3 4+ + + (D) ...

z z z2 3 42 3 4− +

SOL 1.25

1 1 1z z z2 1

21 1< < < <& &

z 11− ...z z z z z

1 1 121 1 1 1 11

2 3= − = + + + +−

b bl l

and z 21− ...z z z z z z

1 1 2 1 1 2 4 81

2 3= − = + + + +−

b bl l

Laurent’s series is given by

( )f z ... ...z z z z z z z z1 1 2 4 98 1 1 1 1 1

2 3 2 3= + + + + − + + + +b bl l

...z z z z1 1 3 7

2 3= + + +b l

& ( )f z ...z z z1 3 72 3 4= + + +


Answer Sheet

1. (A) 6. (A) 11. (B) 16. (C) 21. (D)2. (C) 7. (C) 12. (B) 17. (C) 22. (D)3. (D) 8. (D) 13. (D) 18. (B) 23. (C)4. (B) 9. (A) 14. (C) 19. (A) 24. (B)5. (A) 10. (C) 15. (C) 20. (D) 25. (C)

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Engineering Mathematics GATE